How do I assign a variable a specific scalar value from a matrix?

Hey guys so I'm trying to run this program in which I have two calculated values based on a range of values that the user inputs, and I want to isolate the number which will yield the difference of the two calculated values to be zero. Now I'm not a professional at Matlab so excuse my coding, but this is what it looks like:

Mcr1=0;
Mcr2=0;
y=1.4;
x=0;
Mcr1=input('Enter a minimum value for the Critical Mach Number: ');
Mcr2=input('Enter a maximum value for the Critical Mach Number: '); Mcr=[Mcr1:0.0001:Mcr2];
CPcr1=(-0.43)./(sqrt(1-(Mcr).^2));
CPcr2=(2./(y.*(Mcr).^2)).*(((1+((y-1)/2).*(Mcr).^2)/(1+(y-1)/2)).^(y/(y-1))-1);
CPcr=CPcr1-CPcr2;
CP=abs(CPcr);
fprintf('\n')
disp('    Mcr    CPcr1      CPcr2      Difference')
table=[Mcr',CPcr1',CPcr2',CPcr'];
disp(table)
while x<10;
*act_diff=str2num(sprintf('%.4f',CP));
if act_diff<=0.0001
  fprintf('The Mach number %.4f is the critical Mach number',Mcr)*
  break
else
  x=menu('There was no definite value of the critical Mach number. Would you like to calculate again with different values?','Yes','No');
  switch x
      case 1
          continue
      case 2
          break
  end
end
end

Now the part of the code in which I'm experiencing difficulties is within the asterisks.

fprintf('The Mach number %.4f is the critical Mach number',Mcr)

However Mcr is a vector, but I only need the SINGLE number (or even the two or three, however in that case my fprintf command should also be changed I think) in which the difference is as specified (less than 0.0001)

How do I get Matlab to isolate the value of Mcr from the range that has been input for which CP is less than 0.0001?

Thanks for the help!

Answer 1

[Star Strider]

I just ran your code from the ‘Mcr’ assignment to the ‘CP’ assignment. I have no idea what values ‘Mcr1’ and ‘Mcr2’ are supposed to be, so I just made some up to test it.

If you want the minimum value of ‘CP’, the min function — with two outputs — will give you the minimum value and its index in the vector, so you can use the index to get the corresponding ‘Mcr’ value:

[CPmin,idx] = min(CP);

You might want to put this just after the ‘CP’ assignment. It could make your subsequent code more efficient and easier to work with.

By the way, if you are using R2013b or later, consider changing the name of your ‘table’ variable. The table function creates a table (something you might want to do, actually), and naming it as a variable will cause problems if you want to use the table function later in your code.

Also, consider replacing your input statements with inputdlg. It keeps Command Window clutter to a minimum. Note that the assignments from inputdlg are cells, so you have to add a line to convert them to double, but that’s relatively trivial.

[mentalfiction]

Thank you so much for answering! I will edit the code with what you've suggested and see if that works. I'm using R2012a so I don't think it matters.

Also, I'm sorry if this is supposed to be very obvious, but how do I use the index to get the corresponding Mcr value?

[Star Strider]

My pleasure!

To get the corresponding ‘Mcr’ value, you would assign it as:

q = Mcr(idx);

I used ‘q’ as a default assignment, so change it to a variable that makes sense in the context of your code.

[mentalfiction]

I realized if I use the minimum value of CP, I don't need to enter a while loop. I've edited the program with what you suggested and it works perfect. Thank you so much once again!

Mcrmin=0;
Mcrmax=0;
gamma=1.4;
Mcrmin=input('Enter a minimum value for the Critical Mach Number: ');
Mcrmax=input('Enter a maximum value for the Critical Mach Number: ');
Range=[Mcrmin:0.0001:Mcrmax];
CPcr1=(-0.43)./(sqrt(1-(Range).^2));
CPcr2=(2./(gamma.*(Range).^2)).*(((1+((gamma-1)/2).*(Range).^2)/(1+(gamma-1)/2)).^(gamma/(gamma-1))-1);
Diff=CPcr1-CPcr2;
CP=abs(Diff);
[CPmin,idx] = min(CP);
Mcr = Range(idx);
fprintf('\n')
disp('    Mcr    CPcr1      CPcr2      Difference')
table=[Range',CPcr1',CPcr2',Diff'];
disp(table)
fprintf('The Mach number %.4f is the critical Mach number',Mcr)

[Star Strider]

My pleasure!

Answer 2

[Image Analyst]

In your while loop, what changes? Nothing as far as I can see. You should rename "x" to buttonNumber because that's what it is. People will initially think it has something to do with your y value, but it doesn't. Anyway, in your while loop you just keep looping without ever changing CP or Mcr - why????? Why do you think anything will ever change that will let you break out of the loop by causing act_diff<=0.0001????

And of course Mcr is a vector - that's what you programmed up so I not sure what you're expecting or asking when you say "I only need the SINGLE number (or even the two or three". I don't know what your plans for it are but you can just index it like Star showed you to pull out 1, 2, or 3 values for Mcr.

[mentalfiction]

Oh you're right! I should put the input into the while loop. I didn't notice that because I was focusing on fixing the other issue. Thanks for pointing it out.

And what I was trying to find was the value of Mcr for which CPcr1 = CPcr2 (which will be when their difference = 0 or at least approaches zero).

As for the naming the variables, this is just a rough version, so once I can fix the problems I'm facing I'll rename all the variables to make it more cohesive.

Thanks for answering!